27.6 Runge’s Theorem

Proof, [Proposition VIII.1.1, Con78]. Via fattening, let $V \in \cn_{\complex}^{o}(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^{2}$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)}\subset U$ such that:

1. (1)

   For each $1 \le j \le n$, $R_{j} = [x_{j}, x_{j} + \delta] \times [y_{j}, y_{j} + \delta] \subset U$.

2. (2)

   $\bigcup_{j = 1}^{n} R_{j} \supset V$.

3. (3)

   $x_{j}, y_{j} = 0 \mod \delta$.

In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let

and $\gamma_{j} = \gamma_{j, \downarrow}\cdot \gamma_{j, \leftarrow}\cdot \gamma_{j, \uparrow}\cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_{j}$,

by Cauchy’s Integral Formula. Let $\seqf[N]{\mu_j}$ be an enumeration of

then for each $z_{0} \in \bigcup_{j = 1}^{n} R_{j}^{o}$ and $f \in H(U; E)$,

From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_{j}, \mu_{k})$ is redundant if for each $t \in [0, 1]$, $\mu_{j}(t) = \mu_{k}(1 - t)$. If $(\mu_{j}, \mu_{k})$ are redundant, then

1. (4)

   There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_{j}, \mu_{l})$ or $(\mu_{k}, \mu_{l})$ is redundant.

2. (5)

   $\int_{\mu_j}\frac{f(z)}{z - z_{0}}dz + \int_{\mu_k}\frac{f(z)}{z - z_{0}}dz = 0$.

so every line segment either cannot form a redundant pair, or forms a unique one.

By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_{j}, \mu_{k})$ is redundant. By (5), for each $z_{0} \in \bigcup_{j = 1}^{n} R_{j}^{o}$ and $f \in H(U; E)$,

Let $1 \le j \le N$ such that $\mu_{j}([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_{\complex}(K)$, there exists $1 \le k \le N$ such that $(\mu_{j}, \mu_{k})$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^{m} \mu_{j}([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^{n} R_{j}^{o}$ is dense in $K$, the above also holds for all $z \in K$.

Finally, let $1 \le j \le m$. Since $\mu_{j}$ does not form a redundant pair, $\mu_{j}(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_{k}(0) = \mu_{j}(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves.$\square$

Proof, [Lemma VIII.1.5, Con78]. Since the mapping

is continuous, it is uniformly continuous by Proposition 6.4.5. Hence the mappings $\bracs{\varphi(z_0, \cdot)|t \in [a, b]}$ are uniformly equicontinuous. Thus there exists $\delta > 0$ such that for each $s, t \in [a, b]$ with $|s - t| \le \delta$,

for all $z_{0} \in K$.

Let $(P = \seqfz{t_j}) \in \scp([a, b])$ with $\sigma(t) < \delta$, and

then $R \in \complex(z) \cap H(\complex \setminus \gamma([a, b]); \complex)$, and for each $z_{0} \in K$,

$\square$

Proof, [Lemma III.1.10, Con78]. First suppose that $\infty \not\in P$. Let $A \subset \complex_{\infty} \setminus K$ be the collection of elements such that the lemma holds^[1].

Let $a_{0} \in \complex \setminus K$, then there exists $V \in \cn_{\complex}(a)$ compact such that $V \cap K = \emptyset$. In which case, the function

is continuous, and hence uniformly continuous by Proposition 6.4.5. Thus $1/(z - a) \to 1/(z - a_{0})$ uniformly on compact sets as $a \to a_{0}$, and if $a_{0} \in \ol{A}$, then $a_{0} \in A$ as well. Therefore $A$ is a closed subset of $\complex_{\infty} \setminus K$.

Now, let $a_{0} \in A \cap \complex$, then by Proposition 8.2.5, $r = d(a_{0}, K) > 0$. Let $a \in B_{\complex}(a_{0}, r)$, then for each $z \in K$,

Since $\sup_{z \in K}|a - a_{0}|/|z - a_{0}| < 1$,

where the convergence is uniform on $K$. Therefore $B_{\complex}(a_{0}, r) \subset A$ as well.

Finally, for each $a \in \complex \setminus K$ with $|a| > \sup_{z \in K}|z|$,

where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$.

Since $P \cup \bracs{\infty}\subset A$, and $A$ is an open and closed subset of $C_{\infty} \setminus K$, $A$ is a union of connected components of $\complex_{\infty} \setminus K$ by Lemma 5.12.5. Given that $P$ intersects every connected component of $\complex_{\infty} \setminus K$, the lemma holds for all $a \in \complex \setminus K$.$\square$

Proof. Let $U \in \cn_{\complex}(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By Proposition 27.6.1, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_{0} \in K$,

Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by Lemma 27.6.2, there exists $\seqf{R_j}\subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_{0} \in K$ and $1 \le j \le n$,

so

for all $z_{0} \in K$. By the pole pushing lemma, there exists $S \in \complex(z) \cap H(\complex_{\infty} \setminus P)$ such that $|S(z_{0}) - \sum_{j = 1}^{n} R_{j}(z_{0})| < \eps$ for all $z_{0} \in K$.$\square$